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The main cable of a suspension bridge forms a parabola, described by the equation [tex]y=a(x-h)^2+k[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & 0 & 52.5 & 105 & 157.6 & 210 \\
\hline
$y$ & 27 & 12 & 7 & 12 & 27 \\
\hline
\end{tabular}
\][/tex]

- [tex]y[/tex] = height in feet of the cable above the roadway
- [tex]x[/tex] = horizontal distance in feet from the left bridge support
- [tex]a[/tex] = a constant
- [tex]\((h, k)\)[/tex] = vertex of the parabola

What is the vertex of the parabola?

[tex]\[
\boxed{}
\][/tex]


Sagot :

To solve for the vertex of the parabola given by the equation [tex]\( y = a(x-h)^2 + k \)[/tex] with the provided points, follow these steps:

1. Identify Symmetry:
A parabola exhibits symmetry around its vertex. The points with the same y-value on either side of the vertex can help identify the x-coordinate of the vertex. Specifically, the midpoint of the x-values of the points equidistant from the vertex will give the x-coordinate of the vertex.

2. Given Points:
The points provided are:
[tex]\[ \begin{array}{cc} (0, 27), & (52.5, 12), & (105, 7), & (157.6, 12), & (210, 27) \\ \end{array} \][/tex]

3. Calculate the x-coordinate (h):
Look at the points with the same y-values on either side:
- From [tex]\((0, 27)\)[/tex] and [tex]\((210, 27)\)[/tex]

Compute the midpoint of the x-values:
[tex]\[ h = \frac{0 + 210}{2} = 105 \][/tex]

4. Identify the y-coordinate (k):
At [tex]\( x = 105 \)[/tex], the corresponding y-value is given as:
[tex]\[ k = 7 \][/tex]

Thus, the vertex of the parabolic curve is [tex]\((105, 7)\)[/tex].
[tex]\(\boxed{105}\)[/tex]
[tex]\(\boxed{7}\)[/tex]